It is well known that $$\operatorname{Ai}(z)=\sum _{k=0}^{\infty } \frac{z^{3 k}}{3^{2 k+\frac{2}{3}} k! \Gamma \left(k+\frac{2}{3}\right)} -\sum _{k=0}^{\infty } \frac{z^{3 k+1}}{3^{2 k+\frac{4}{3}} k! \Gamma \left(k+\frac{4}{3}\right)}.$$
What are the steps to convert the above expression to the more concise form (https://mathworld.wolfram.com/AiryFunctions.html)
$$\operatorname{Ai}(z)=\frac{1}{\pi 3^{2/3}}\sum _{k=0}^{\infty } \frac{ \Gamma \big(\frac{k+1}{3}\big) \sin \left(\frac{1}{3} (2 \pi ) (k+1)\right) \left(\sqrt[3]{3} z\right)^k}{k!} .$$